The function
$f(x) = \sum_{1\leq n \leq x} \lfloor \frac x n \rfloor ^{-\theta} $
can be expressed as
$$f(x) = \sum_k k^{-\theta} \cdot\#\{n : \lfloor \frac x n \rfloor = k\}$$
$$ = \sum_k k^{-\theta} \left(\lfloor \frac x k\rfloor - \lfloor \frac x {k+1}\rfloor \right)$$
$$ = \sum_k k^{-\theta}\left( \frac{x}{k^2+k} + O(1)\right)$$
so  the constant 
$c(\theta)= \displaystyle\sum_{k\geq 1} \frac{k^{-\theta}}{k^2 + k}$.

To get a better bound on the error term, we can regroup terms
$$f(x) = \lfloor x\rfloor - \sum_{k\geq2} ((k-1)^{-\theta}-k^{-\theta})\lfloor \frac{x}{k} \rfloor$$
so the error from ignoring the ''floor'' signs is
$$ |c(\theta)x - f(x)| = \left| (x-\lfloor x\rfloor) - \sum_{k\geq2} ((k-1)^{-\theta}-k^{-\theta})\left( \frac{x}{k} - \lfloor \frac{x}{k}\rfloor\right) \right|$$
$$\leq 1 + \sum_{k \geq 2}((k-1)^{-\theta}-k^{-\theta}) = 2.$$